TITLE: Convergence in probability of the product of two random variables
QUESTION [30 upvotes]: Suppose $\{X_n\}$ and $\{Y_n\}$ converge in probability to $X$ and $Y$, respectively.
Will $X_n Y_n$ converge in probability to $X Y$?
I know the answer is yes.
If we treat $(X_n,Y_n)$ as a random vector, and it converges in probability to $(X,Y)$ by the assumption. Then $g(x,y) = xy$ is a continuous function and according to the continuous mapping theorem, $g(X_n,Y_n)$ converges in probability to $g(X,Y)$.
My question is how to go from the definition without using the continuous mapping theorem.
My attempt is as follows.
$$P(|X_nY_n-XY|>\epsilon)=P(|X_nY_n-X_nY+X_nY-XY|>\epsilon)$$
$$\leq P(|X_n(Y_n-Y)|+|Y(X_n-X)|>\epsilon)$$
It seems tempting to conclude that the last term goes to zero as $n$ goes to infinity.
But I am not sure about it.
Am I right or did I miss something?

REPLY [3 votes]: First,
\begin{align}
|X_nY_n-XY| &\le |(X_n-X)(Y_n-Y)|+|(X_n-X)Y|+|(Y_n-Y)X| \\
&\le \frac{1}{2}(X_n-X)^2+\frac{1}{2}(Y_n-Y)^2+|(X_n-X)Y|+|(Y_n-Y)X|.
\end{align}
Then for any $\epsilon>0$ and $K>0$
\begin{align}
P\{|X_nY_n-XY|>\epsilon\} &\le P\{|X_n-X|>\sqrt{\epsilon/2}\}+ P\{|Y_n-Y|>\sqrt{\epsilon/2}\} \\
&+P\{|X_n-X|>\epsilon/(4K)\}+P\{|Y_n-Y|>\epsilon/(4K)\} \\
&+P\{|X|>K\}+P\{|Y|>K\}; \\
\\
\because \{|(X_n-X)Y|>\epsilon/4\}&\subset\{|Y|>K\}\cup \{|X_n-X|>\epsilon/(4K)\},\\
\{|(Y_n-Y)X|>\epsilon/4\}&\subset\{|X|>K\}\cup \{|Y_n-Y|>\epsilon/(4K)\}.
\end{align}
For any $\nu>0$ we can find $K$ s.t. $P\{|X|>K\}<\nu$ and $P\{|Y|>K\}<\nu$ so that
$$\limsup_{n\to\infty}P\{|X_nY_n-XY|>\epsilon\}\le 2\nu.$$
Now, send $\nu\downarrow 0$.